There are exactly three integers $x$ satisfying the inequality
\[x^2 + bx + 2 \le 0.\]How many integer values of $b$ are possible?
Explanation: The roots of the corresponding equation $x^2 + bx + 2 = 0$ are
\[\frac{-b \pm \sqrt{b^2 - 8}}{2}.\](Note that these roots must be real, otherwise, the inequality $x^2 + bx + 2 \le 0$ has no real solutions.)  Thus, the solution to this inequality $x^2 + bx + 2 \le 0$ is
\[\frac{-b - \sqrt{b^2 - 8}}{2} \le x \le \frac{-b + \sqrt{b^2 - 8}}{2}.\]If the length of this interval is at least 4, then it must contain at least 4 integers, so the width of this interval must be less than 4.  Thus,
\[\sqrt{b^2 - 8} < 4.\]Then $b^2 - 8 < 16,$ so $b^2 < 24.$  We must also have $b^2 > 8.$  The only possible values of $b$ are then $-4,$ $-3,$ 3, and 4.  We can look at each case.

\[
\begin{array}{c|c}
b & \text{Integer solutions to $x^2 + bx + 2 \le 0$} \\ \hline
-4 & 1, 2, 3 \\
-3 & 1, 2 \\
3 & -2, -1 \\
4 & -3, -2, -1
\end{array}
\]Thus, there are $\boxed{2}$ values of $b$ that work, namely $-4$ and 4.